Long answer: By "no mass" you probably mean "no rest mass". Your question is almost a FAQ in these forums, I'm sure if you search you'll find loads of relevant threads (I recall in particular this recent one).
[edit]Also this and this thread may be useful.[/edit]

[edit (2)]I didn't know the word compfuzzled, but I like it. Does it actually exist?[/edit]

I don't think kinetic energy is the correct word, it's nothing like [itex]\frac12 m v^2[/itex] (of course, because there is a mass in there). The energy of a photon depends on its wave length. I added two more threads in an edit of my first post.

Yes and no. Most of us prefer to write [itex]E^2=\vec p^2c^2+m^2c^4[/itex], where E is the energy, p is the momentum and m is the mass. Some people would write m0 instead of m on the right-hand side, and define m (yes, define is the right word) by saying that this is equal to [itex]m^2c^4[/itex]. They would call m the "mass" and m0 the "rest mass".

[itex]E=mc^2[/itex] is often quoted as Einstein's energy equation, but it isn't really the full equation. The full relationship for the energy of a particle is given by:

[tex]E^2 = \left(m_0c^2\right)^2 + \left(pc\right)^2[/tex]

Where [itex]m_0[/itex] is the invariant or "rest" mass of the particle and [itex]p[/itex] is it's momentum. The first term of the expression is sometimes called the "rest energy" of a particle, because that is the energy that the particle has as measured from a reference frame stationary relative to it.

Now, photons have zero mass and therefore the first term disappears, i.e. the photon has no "rest energy". We are now simply left with:

[tex]E = pc[/tex]

So all the photon's energy is in fact due to the photon's motion, as you have said previously.

i realised that wasntthe entire equation. i just shortened it to make it easy. is that taboo around here?lol

No not at all, I just thought that the full equation clarified the explanation somewhat. In fact your equation [itex]E=mc^2[/itex] can be considered complete if one considers [itex]m[/itex] as relativistic mass rather than invariant mass.

Sorry if the question is a bit simple, im not a physicist by profession. I seek to understand how a photon can move if it has 0 mass. Is it that photons aquire some sort of "mass" and thus can move (therefore fulfilling newtons laws)? I understand that we dont know what "mass" really is but I am perplexed as to how a photon moves in spacetime. Any answers would be appreciated.

[itex]E=mc^2[/itex] is often quoted as Einstein's energy equation, but it isn't really the full equation. The full relationship for the energy of a particle is given by:

[tex]E^2 = \left(m_0c^2\right)^2 + \left(pc\right)^2[/tex]

Where [itex]m_0[/itex] is the invariant or "rest" mass of the particle and [itex]p[/itex] is it's momentum. The first term of the expression is sometimes called the "rest energy" of a particle, because that is the energy that the particle has as measured from a reference frame stationary relative to it.

Now, photons have zero mass and therefore the first term disappears, i.e. the photon has no "rest energy". We are now simply left with:

[tex]E = pc[/tex]

So all the photon's energy is in fact due to the photon's motion, as you have said previously.

I seek to understand how a photon can move if it has 0 mass. Is it that photons aquire some sort of "mass" and thus can move (therefore fulfilling newtons laws)? I understand that we dont know what "mass" really is but I am perplexed as to how a photon moves in spacetime.

So you are saying that a photons ability to travel at the speed of light is attributed to their momentum but newtonian physics states p=mv.
If photons have no mass then isnt p=0?

Newton's laws of collisions only require momentum.

Newton's laws of collisions are completely valid. They accord with reality.

Newtonian physics of space-time is wrong (or, if you prefer, inaccurate) … there is no point in trying to apply it to objects moving at or near the speed of light.

Newtonian physics of space-time states p=mv. That is wrong (but accurate enough for speeds nowhere near the speed of light). There is no point in asking how a photon can fulfil a law that is wrong: it can't and it doesn't!

The correct law for momentum is p = mv/√(1 - v2/c2) …

if m = 0, p doesn't have to be zero if v/√(1 - v2/c2) is infinite, ie if v = c.

So you are saying that a photons ability to travel at the speed of light is attributed to their momentum but newtonian physics states p=mv. If photons have no mass then isnt p=0?

Newtonian physics is not exactly correct. The momentum of a massive particle in special relativity is

[tex]p=\frac{mv}{\sqrt{1-v^2/c^2}}[/tex]

if you make the velocity (v) get closer to c and keep the mass constant, then the term in the denominator gets smaller and smaller and the momentum gets larger and larger. If you know that the momentum when v=c is finite, then as v gets closer to c, you must have the mass get smaller and smaller. You can see that only a massless particle can travel at the speed of light without having infinite momentum. The problem is that when v=c you have for momentum 0/0 which is undefined. Newtonian mechanics is WAY wrong at or near the speed of light, so you cannot use Newtonian definitions to figure out what is happening at or near the speed of light.